Tables, Graphs and Numerical Summaries of Data Assignment Task 1 In Task 1 you are required to use Excel to produce a frequency Column Chart and a relative frequency Pie-Chart for your sample to show...

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Answered Same DayDec 20, 2021

Answer To: Tables, Graphs and Numerical Summaries of Data Assignment Task 1 In Task 1 you are required to use...

David answered on Dec 20 2021
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1. The following figure shows the frequency column chart for Building Type:
The following figure shows the pie chart for Building Type:


(a) 20 properties in my sample consist of Brick Veneer Building.
(b) Brick Veneer Building Type occurs most frequently in my sample.
(c) Proportion of properties in my sample consists of vacant land = 2/50 = 0.04
0
5
10
15
20
25
Br BrV Wbrd VL
Fr
e
q
u
e
n
cy

Building Type
Frequency Column Chart for Building type
Br
34%
BrV
40%
Wbrd
22%
VL
4%
Pie Chart for Building Type
Bld.Type Frequency
Br 17
BrV 20
Wbrd 11
VL 2
2.
Advertised price($000) Frequency


0-<100 1
100-<200 3
200-<300 6
300-<400 11
400-<500 3
500-<600 8
600-<700 8
700-<800 5
800-<900 4
900-<1000 0
1000-<1100 1
(a) The data has a tendency to cluster in the interval 300-<400.
(b) The dispersion of the data is high as the spread is almost over the entire range. (need to
calculate the standard deviation to show dispersion)
[I am quoting here the exact question from the assignment – it clearly states without
generating any numerical summary then why should one calculate std deviation? –
Task 2 (5 marks)
Use Excel to produce a Frequency Histogram for the “Advertised Price” variable in your sample.
Without generating numerical summary measures for this data, use this histogram to describe,
(a) Any tendency for the data to cluster around a particular value or group of values.
(b) The dispersion of the data.
(c) The shape of the data.]
(c) The data seems to be bi-modal in shape (M-shape).
3. (a)
112 385 453 710 880
190 401 466 710 900
209 409 467 715 900
209 410.5 583.5 718 960
260 420.5 590 720 1180
270 427 600 750
331 430 605 835
340 450 615 840
347 450 650 845
363 452.5 704.5 850
(b) We know that p-th percentile is given by the percentile location formula: LP = (n+1)P/100.
The first quartile,Q1 = 25-th percentile,L25 and the third quartile, Q3 = 75-th percentile, L75
0
2
4
6
8
10
12
Fr
e
q
u
e
n
cy

Advertised Price($000)
Frequency Histogram for Advertised Price
Therefore, 60-th percentile, L60 = 27-th Observation = 605
The first quartile, L25 = 11.25 ≈ 12-th Observation = 401
The third quartile, L75 = 33.75 ≈ 34-th Observation = 718
(c)Since L60 is 605 then 60% of the sample sold prices are less than equal to L60, i.e., 605 .
(d) The inter quartile range = L75 – L25 = 718 – 401 = 317 which implies that 50% of the sample sold prices
are between 718 and 401 and it has a spread of 317.
(Need to give the explanation of part b and c)
4. (a) Table showing the Descriptive Statistics for Sold price($000):
Sold price($000)
Mean 558.078
Standard Error 35.907
Median 467.000
Mode 209.000
Std Deviation 240.873
Sample Variance 58019.931
Kurtosis -0.432
Skewness 0.340
Range 1068.000
Minimum 112.000
Maximum 1180.000
Sum 25113.500
Count 45
(b) Using the value of Q1, Q3 and IQR from 3.(b) and 3.(d) we have Q1 – 1.5 x IQR = -74.5 and Q3 + 1.5 x
IQR = 1193.5 (what is the value of Q1 and Q3)
(c) These two limits are used for outlier detection – any observation less than the lower limit or greater
than the upper limit is an outlier. Clearly from the above table we see that the maximum value is =1180
and minimum value is = 112. Therefore, there is no outlier in the sample. Thus we can use Mean as an
appropriate measure of central tendency (since mean is sensitive to outliers but there is no outlier in the
sample). An appropriate measure of dispersion is Standard Deviation since there is no outlier in the data
it shows that the range of the sample is not unusually high. Hence we can use Standard Deviation.
5. (a)...
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